JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
Resistance of the wire is measured as 2 Ω and 3 Ω at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is :
- (a) 0.033 °C⁻¹
- (b) -0.033 °C⁻¹
- (c) 0.011 °C⁻¹
- (d) 0.055 °C⁻¹
Answer and solution
Answer (a) 0.033 °C⁻¹
Solution
Physics sir · 4 steps
Write R = R₀(1 + α t) at both temperatures and divide, so R₀ at 0°C cancels out.
- Set up equations. The temperature coefficient is defined with respect to the resistance R₀ at 0°C. So 2 = R₀(1 + 10α) and 3 = R₀(1 + 30α).
3/2 = (1 + 30α)/(1 + 10α)
- Solve for alpha. Cross-multiply to get 3 + 30α = 2 + 60α, so 30α = 1 and α = 1/30 ≈ 0.033 °C⁻¹. As a check, R₀ = 1.5 Ω and 1.5(1 + 30 × 1/30) = 3 Ω.
- Sign check. Resistance rises from 2 Ω to 3 Ω as the wire heats up, so α must be positive, as it is for a metal.
- So. The temperature coefficient is 0.033 °C⁻¹, which is option (a).
The idea to keep. When you are given resistances at two temperatures, write R = R₀(1 + α t) for each and take the ratio. R₀ cancels and α comes out directly. Resistance going up means α is positive.
Answer from the NTA final answer key. Solution by PrepWiser.